Context: Given is the scalar system $$\dot{x}=-ax+u$$ with $x(0)=x_0$ and $|u(t)|\leq M$ for a constant $a$ and a positive constant $M$ along with cost criterion $$J(u)=x(1)+\frac{1}{2}\int_{0}^{1}u^2(t)dt.$$
Tasks:
Determine the Hamiltonian for the optimal control problem and the differential equation for the costate $p$.
Determine the optimal input in the case that $M=\infty$.
Determine the optimal input in the case that $M$ is finite, distinguishing between the cases $a\geq0$ and $a<0$.
My attempt: For the first task, I found the Hamiltonian $$H(x,p,u)=p(-ax+u)+\frac{u^2}{2}$$ and that the differential equation for the costate is given by $$\dot{p}(t)=ap(t)$$ with the conditions $p(1)=1$, the solution of which is $$p(t)=e^{a(t-1)}.$$
For the second task, in minimising the Hamiltonian, we find that the optimal input is given by $$u(t)=-p(t)=-e^{a(t-1)}.$$
Question: From here my question is what effect the bound $|u(t)|\leq M$ has on this optimal input. Is it maybe that for particular values of finite $M$, the optimal input $u(t)$ is not given by $-e^{a(t-1)}$? Any help is appreciated.
The optimal control in general when using Pontryagin's maximum principle can be found by minimizing the Hamiltonian with respect to the control variables. In your case this would be the constrained optimization problem
\begin{align} \min_{u}\ & p(-ax+u)+\frac{u^2}{2} \\ \text{s.t.}\ & |u| \leq M \end{align}
Since the cost function is convex and the decision variable is scalar the solution would either be equal to the unconstrained solution ($u(t)=-p(t)$) or on the closest boundary of the constraint if that solution violates that constraint. Therefore, the constrained solution can be written as
$$ u(t) = \left\{ \begin{array}{ll} M & \text{if}\ \, p(t) < -M, \\ -M & \text{if}\ \, p(t) > M, \\ -p(t) & \text{otherwise}. \end{array} \right. \tag{1} $$
Since the solution for $p(t)$ in this case is always positive one could also remove the expression for $p(t) < -M$ from $(1)$. Depending on the values of $M$ and $a$ only one or two expressions from $(1)$ will be used. For example for $M<1$ the expression from $p(t) > M$ will always be used, since $p(1)=1$, but for sufficiently large $a$ it would at least be the case that $p(0) \ngtr M$. It can be noted that the exponential expression for $p(t)$ is monotonicly increasing or decreasing, so there will be at most one switch of which expression is being used from the right hand side from $(1)$. Furthermore, in the limit of $M$ to infinity (and finite $a$) the constraint problem becomes equivalent to the unconstrained case.